So one "case" has to be, when the balls are considered (at least logically) numbered from $1-9$, whith $1-5$ being green and $6-9$ being white: And remembering the $997$ green vs $3$ white balls example, you cannot forget that all those $997$ green balls are different and you cannot just "combine" them, that will give an incorrect result. So, the assumption you should be using is that each individual draw of $3$ balls is equally likely. That's a good model for the real world, ifĪ) the balls aren't distinguishable by anything for the "drawing process" (be that a person grabbing into it or some machine doing it, like for many lotteries),ī) the balls have been sufficiently mixed initially.Įspecially a) is tricker than it sounds, if you want to prevent cheating. The unspoken assumption on all those "draw out of a bag" questions is that each individual "draw" is getting each remaining ball with equal probability. How to decide if the cases are actually equally probable is something between math and the actual real world events that you try to model. You can get to absurd conclusions if you forget that, like that there are just two possible outcomes to the event "Then sun will go up sometime in NY City in the next 48h.", so the probability of that event happening (for each possible "now") is just $\frac12$, right? So again, the fundamental requirement for "Favorable cases / Total cases" to work is that the cases are actually all equally probable. If you think they don't, (which is the correct intuition most people have in that very lopsided example), there is then also no reason to think they are exactly equal in the more "even" case of $5$ green and $4$ white balls in the bag.Īnd that doesn't say anything about how they relate to the other cases of " $2$ green and $1$ white ball drawn", and the color reversed. Would you think that drawing $3$ green and $0$ white balls has the same probability as drawing $0$ green and $3$ white balls? Let's consider a different problem, we have $1000$ balls in that bag, $997$ green balls and $3$ white balls to choose from. So let's first see why your $4$ cases ( $3$G + $0$W, a.s.o) are not equally likely. That's the scenario where most introduction examples start, but that side condition is not usually stressed in teaching, and (if mentioned at all) usually soon forgotten by the students, because it seems "obvious" when mentioned, but can have subtle consequences, as you experienced here. The main reason your argument is incorrect is that the formula "Favorable cases / Total cases" only works if each "case" is as probable as each other case. ![]() That's a very common problem to have, it crops up here regularly from time to time and can also "trick" people that are more experienced than you in statistics/probability theory.
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